Let A, B, C be events. Show that if P(A|B) = 1, then P(Bc|Ac) = 1 Bc=
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Question:
Let A, B, C be events.
Show that if P(A|B) = 1, then P(Bc|Ac) = 1
Bc= B complement, Ac = A complement.
I got:
P(A|B) = P(A and B)/ P(B)
1= P(A)/P(B) because P(A|B)=1
1-P(A) = 1-P(B)
P(Ac) = P(Bc)
P(Bc)/P(Ac) = 1 and this is only true if P(Bc|Ac) = 1
Is this right?
Show that if P(A|B) = 1, then P(Bc|Ac) = 1
Bc= B complement, Ac = A complement.
I got:
P(A|B) = P(A and B)/ P(B)
1= P(A)/P(B) because P(A|B)=1
1-P(A) = 1-P(B)
P(Ac) = P(Bc)
P(Bc)/P(Ac) = 1 and this is only true if P(Bc|Ac) = 1
Is this right?
Related Book For
Discrete and Combinatorial Mathematics An Applied Introduction
ISBN: 978-0201726343
5th edition
Authors: Ralph P. Grimaldi
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